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Statistics

Notes Mean of t

The mean of t is defined as its expected value, given by

¥ 1
E ( ) t = ò t . .e - mt dt = , where m denotes the average number of occurrence of events per
m
0 m
unit of time or distance.

Example 24: A telephone operator attends on an average 150 telephone calls per hour.
Assuming that the distribution of time between consecutive calls follows an exponential
distribution, find the probability that (i) the time between two consecutive calls is less than 2
minutes, (ii) the next call will be received only after 3 minutes.
Solution.

150
Here m = the average number of calls per minute = = 2.5.
60

2 - 2.5t
(i) P (t £ ) 2 = ò 0 2.5e dt = F ( ) 2

We know that F(t) = 1 - e , \ F(2) = 1 - e -2.5 × 2 = 0.9933
-mt
(ii) P(t > 3) = 1 - P(t £ 3) = 1 - F(3)

= 1 - [1 - e -2.5 × 3 ] = 0.0006

Example 25: The average number of accidents in an industry during a year is estimated
to be 5. If the distribution of time between two consecutive accidents is known to be exponential,
find the probability that there will be no accidents during the next two months.
Solution.
5
Here m denotes the average number of accidents per month = .
12
5
- ´ 2
 P(t > 2) = 1 - F(2) = e 12 = e - 0.833 = 0.4347.

Example 26: The distribution of life, in hours, of a bulb is known to be exponential with
mean life of 600 hours. What is the probability that (i) it will not last more than 500 hours, (ii) it
will last more than 700 hours?
Solution.

1
Since the random variable denote hours, therefore m =
600

1
- ´ 500 - 0.833
-
(i) P(t £ 500) = F(500) 1 e= - 600 = 1 e = 0.5653.
700
- - 1.1667
(ii) P(t > 700) = 1 - F(700) e= 600 = e = 0.3114.

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